Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: PS - Rightmost nonzero digit & product of factors [#permalink]
08 May 2013, 02:44

sergbov123 wrote:

Q1) What is the rightmost non-zero digit of 20!? a) 2 b) 3 c) 5 d) 6 e) 8

Q2) What is the product of the factors of 432? a) 2^90 b) 3^78 c) 2^40 * 3^30 d) 2^25 * 3^75 e) 2^5 * 3*10 * (2^25 * 3^30 - 1)

The easiest way to solve Q2 is as follows:

There is a formula to find the product of factors of a number 'n'. => (Prime factorization of n) ^ f/2 , where f is the total no. of factors.

Coming back to the quqestion, prime factorization of 432 = 2^4 * 3^3 Hence, total no.of factors = (4+1)*(3+1) = 5*4 = 20 The product of all factors of 432 = (2^4 * 3^3)^(20/2) = (2^4 * 3^3)^10 = 2^40 * 3^30 = C

Re: PS - Rightmost nonzero digit & product of factors [#permalink]
08 May 2013, 05:15

holy wrote:

shrouded1 wrote:

mainhoon wrote:

What is the answer for 20!'s last digit?

Using the same approach I showed for 90!

Set 1 : {2,12} Set 2 : {5,15} Set 3 : {10,20} Set 4 : {1,3,4,6,7,8,9, 11,13,14,16,17,18,19}

This reduces to (canceling 2s&5s in set 1 and set 2 + canceling 10s in set 3) :

Set 1 : {1,6} --> Last digit 6 Set 2 : {1,3} --> Last digit 3 Set 3 : {1,2} --> Last digit 2 Set 4 : {1,3,4,6,7,8,9, 11,13,14,16,17,18,19} --> Last digit 8*8=4

So overall last digit = 6*3*2*4 = 4

Hence last non-zero digit of 20! = 4

Can you please explain how you calculating the last digit for Set 4 ?

Set 4 is everything else {1,3,4,6,7,8,9,11,13,14,16,17,18,19, 21,23,24,26,27,28,29.........81,83,84,86,87,88,89} notice the repeating pattern is set 4 - 1,3,4,6,7,8,9,1,3,4,6,7,8,9,........ and there are 9 such patterns (1,3,4,6,7,8,9) (11,13,14,16,17,18,19).....(81,83,84,86,87,88,89)

Multiply the digits in the patter 1*3*4*6*7*8*9 - the units digit is 8

Quote:

The "cyclicity" of 8 in a product is 8^1=8 8^2=64 8^3=512 8^4=4096 8^5=32768 and so on, the pattern is 8,4,2,6....

And as the pattern (1,3,4,6,7,8,9) repeats 9 times - so the units digit will be 8 _________________

Consider giving +1 Kudo when my post helps you. Also, Good Questions deserve Kudos..!

Re: PS - Rightmost nonzero digit & product of factors [#permalink]
09 May 2013, 04:48

shrouded1 wrote:

gurpreetsingh wrote:

Bump !!

There is a brute force way to solve this. The principle is the same, you have to remove all powers of 5 and a corresponding number of powers of 2 from the product. The product of the rest will give you the non-zero digit.

First of all what I'd do is split the numbers out into four sets. Set 1 is {2,12,22, ... ,82}. Set 2 is {5,15,25,...,85}. Set 3 is {10,20,30,...90} and Set 4 is everything else {1,3,4,6,7,8,9, 11,13,14,16,17,18,19, .... ,81,83,84,86,87,88,89}

Set 1,2,3 have 9 elements each and have all the powers of 5 within them. Set 4 has the remaining 7x9=63 elements

Lets process Set 3 first, I can take out all powers of 10, to get {1,2,3,4,5,6,7,8,9} Then I take out the 5 and a corresponding 2 to get {1,1,3,4,1,6,7,8,9}

Now lets process Set 1 & 2, For every 5 I take out of set 2, I will take out a 2 from set 1 : So set 1 becomes : {1,6,11,16,....,36,41} and set 2 : {1,3,5,7,9,11,13,15,17}

We are almost done now, the only powers of 5 left are in set 2 which I take out and against that I take out 4 in set 3 to leave me with with :

Set 1 : {1,6,.....,36,41} : Last digit is 6 Set 2 : {1,3,1,7,9,11,13,3,17} : Last digit is 7 Set 3 : {1,1,3,1,1,6,7,8,9} : Last digit 2 Set 4 : Which has a pattern to it and easy to show last digit is 8 (It has subsets ending in 1,3,4,6,7,8,9 each of which has last digit 8 and 9 such subsets. The "cyclicity" of 8 in a product is 8,4,2,6,repeat so with 9 subsets the answer will be 8) Multiplying those out, the last digit of 90! is 2

The approach is brute force, but once you get a hang of it, it doesnt take very long to calculate it out. I would be very surprised if this is a GMAT level question, unless of course there is a trick to doing this which has completely evaded me

Hi,

I dont understand set 3.. Its ok you took out 2 and 5 , so that they become 1 and 1 respectively. But why 4 in set 3 becomes 1 ..Will you please explain? _________________

Kabilan.K Kudos is a boost to participate actively and contribute more to the forum

Re: PS - Rightmost nonzero digit & product of factors [#permalink]
09 May 2013, 07:02

bhanushalinikhil wrote:

GMAT TIGER wrote:

sergbov123 wrote:

Q2) What is the product of the factors of 432?

a) 2^90 b) 3^78 c) 2^40 * 3^30 d) 2^25 * 3^75 e) 2^5 * 3*10 * (2^25 * 3^30 - 1)

Thats a good & tough question:

a) 2^90 ......... cannot be because it doesnot have any multiple/power of 3. b) 3^78......... cannot be because it doesnot have any multiple/power of 2. c) 2^40 * 3^30......... can be because it has multiples/power of 2 & 3. d) 2^25 * 3^75 ......... can be because it has multiples/power of 2 & 3. e) 2^5 * 3*10 * (2^25 * 3^30 - 1) probably not.

Lets work on c and d. 432 = (2^4) (3^3) Factor of 432 = 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 36, 48, 54, 72, 108, 144, 216 and 432.

Product of 432 = 2x3x4x6x8x9x12x16x18x24x27x36x48x54x72x108x144x216x432 = 2x3x 2^2 x 2x3 x 2^3 x 3^3 x 2^2x3 x 2^4 x 2x3^2 x 2^3x3 x 3^3 x 2^2x3^2 x 2^4x3 x 2x3^3 x 2^3x3^2 x 2^2x3^3 x 2^4x3^2 x 2^3x3^3 x 2^4x3^3 = 2^41 x 3^31

Hmmm ........ I got 2^41 x 3^31 instead of 2^40 x 3^30.

Answer should be C but how it is 2^40 x 3^30? I must have made a mistake somewhere....

Can you find it for me? Thanx....

2x 3x 2^2 x 2x3 x 2^3 x 3^3(p) x 2^2x3 x 2^4 x 2x3^2 x 2^3x3 x 3^3 x 2^2x3^2 x 2^4x3 x 2x3^3 x 2^3x3^2 x 2^2x3^3 x 2^4x3^2 x 2^3x3^3 x 2^4x3^3

3^3(p) x For 9, you have written 3^3 instead of 3^2. But thts all I could find out.

I counted 40 2s, so that is ok...you miscounted... _________________

Re: What is the rightmost non-zero digit of 20!? [#permalink]
27 Jan 2014, 09:06

Question 1:

For the first digit after the 0s, we just multiply the unit digits excluding 0,2,5s 1*3*4*6*7*8*9 = 36288 (if you multiply step by step, drop the whole number and only use the unit digits of your solutions!)

This needs to be done twice, because the unit digits appear between 0 and 10 as well as between 10 and 20.

So you multiply the unit digit from 36288 with 36288: 8*8 = 64 --> First digit after the 0s is 4.

If you want to compute this with 90!, you've got to calculate 8^9 (or step by step with the unit digit of your solutions

64^4*8 -> 4^4 * 8 -> 6^2 * 8 -> 6*8 = 48 Thus the first digits (after the 21) 0s is 8. (back-checked with wolframalpha)

Question 2: (did not use timer, but guess between 1min and 1:30)

432 = 2^4 * 3^3

Attachment:

Screen Shot 2014-01-27 at 18.03.33.png [ 11.21 KiB | Viewed 356 times ]

According to the table, the solution is 2^40 * 3^30 _________________

Thank You = 1 Kudos B.Sc., International Production Engineering and Management M.Sc. mult., European Master in Management Candidate